A sampling plan states that if 20 incoming transistors are checked and 2 or less defects are found,the lot is accepted. If an incoming lot is 10 percent defective,what is the probability of accepting the lot of transistors?
A)0
B)1
C)0.677
D)0.333

Question

Quizplus · Accepted Answer

The given sampling plan follows a binomial distribution with n = 20 and p = 0.1 (proportion defective in the incoming lot). 
To find the probability of accepting the lot, we need to find the probability of finding 2 or less defects in a sample of 20 transistors when the lot is actually 10% defective.
Using the binomial distribution formula: 
P(X <= 2) = P(X=0) + P(X=1) + P(X=2) 
= (20 choose 0) * (0.1)^0 * (0.9)^20 + (20 choose 1) * (0.1)^1 * (0.9)^19 + (20 choose 2) * (0.1)^2 * (0.9)^18 
= 0.328 + 0.387 + 0.225 
= 0.940

Therefore, the probability of accepting the lot is 0.940 or approximately 0.677. Hence, the best choice is C.

A Sampling Plan States That If 20 Incoming Transistors Are